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Chem 1A
Keys to Chapter 4 Exercises
Name:_________________________
Exercises #1:
1.
What types of particles are presents in an aqueous solution of:
(a) Strong electrolytes – ions in high concentration
(b) Weak electrolytes – ions and molecules; the ionic concentration is very low.
(c) Nonelectrolytes – molecules
2.
(a) H2SO4 (aq)
(b) BaCl2(aq)
3.
(a) C2H6O2(aq)
and (d) C12H22O11(aq)
4.
(b) HNO2(aq)
(c) NH3(aq)
5.
(a) NH4Cl: NH4+ & Cl–; strong electrolyte
and
and
(d) NaOH(aq)
(e) HC2H3O2(aq)
(d) HNO3: H3O+ (or simply H+) and NO3–; strong electrolyte
(b) C2H5OH (ethanol): C2H3OH; nonelectrolyte
(c) C6H12O6(glucose): C6H12O6; nonelectrolyte
(f) HC2H3O2: HC2H3O2 (mostly), H3O+, and C2H3O2–; Peak electrolyte
(e) NaOH: Na+ & OH–; strong electrolyte
6.
Ca(NO3)2(aq)  Ca2+(aq) + 2NO3–(aq)
7.
NH4OH(aq) ⇄ NH4+(aq) + OH–(aq)
8.
(b) C2H6O2
9.
15.6%A
10.
(a) 0.1875 mole; (b) 0.2500 M; (c) K2Cr2O7(aq)  2 K+(aq) + Cr2O72–(aq)
(d) 0.7500 M;
(c) C6H12O6 and
(e) C3H7OH
(e) 0.00156 mole.
Exercises #2:
1.
Molecular equation: Pb(NO3)2(aq) + K2CrO4(aq)  PbCrO4(s);
Total ionic equation: Pb2+(aq) + 2NO3–(aq) + 2K+(aq) + CrO42–(aq)  PbCrO4(s) + 2K+(aq) + 2NO3–(aq);
Net ionic equation:
2.
Pb2+(aq) + CrO42–(aq)  PbCrO4(s)
Molecular equation: 3AgNO3(aq) + Na3PO4(aq)  Ag3PO4(s) + 3 NaNO3(aq);
Total ionic equation: 3Ag+(aq) + 3NO3–(aq) + 3Na+(aq) + PO43–(aq)  Ag3PO4(s) + 3Na+(aq) + 3NO3–(aq);
Net ionic equation:
3.
3Ag+(aq) + PO43–(aq)  Ag3PO4(s)
Aqueous calcium hydroxide solution reacts with nitric acid solution to form water and calcium nitrate
solution.
Molecular equation: Ca(OH)2(aq) + 2 HNO3(aq)  2 H2O(l) + Ca(NO3)2(aq);
Total ionic equation: Ca2+(aq) + 2OH–(aq) + 2H+(aq) + 2NO3–(aq)  2H2O (l) + Ca2+(aq) + 2NO3–(aq);
Net ionic equation:
H+(aq) + OH–(aq)  H2O(l)
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Chem 1A
Keys to Chapter 4 Exercises
4.
Name:_________________________
Molecular equation: CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g);
Total ionic equation: CaCO3(s) + 2H+(aq) + 2Cl–(aq)  Ca2+(aq) + 2Cl–(aq) + H2O(l) + CO2(g);
Net ionic equation:
5.
CaCO3(s) + 2H+(aq)  Ca2+(aq) + H2O(l) + CO2(g);
Molecular equation: H2SO4(aq) + Ba(OH)2(aq)  BaSO4(s) + 2H2O(l);
Total ionic equation: 2H+(aq) + SO42–(aq) + Ba2+(aq) + 2 OH–(aq)  BaSO4(s) + 2H2O(l);
Net ionic equation #1:
Ba2+(aq) + SO42–(aq)  BaSO4(s);
Net ionic equation #2:
H+(aq) + OH–(aq)  H2O(l);
Exercises #3:
1.
(a) 0.83 mole
(b) 38 g
(c) 26 gallons
2.
0.150 mole of Al3+ and 0.225 mole of SO42- ions; total mole of ions = 0.375 mole
3.
Concentration of Ba2+ = 0.0520 M
4.
Molarity of HC2H3O2 in vinegar = 0.8215 M
5.
2.40 g of CaC2O4
6.
Percent of Ag = 29.41%; net ionic equation: Ag+(aq) + SCN–(aq)  AgSCN(s)
7.
(a) Balanced equation: Fe(NO3)3(aq) + 3KOH(aq)  Fe(OH)3(s) + 3KNO3(aq);
Mass of Fe(OH)3 produced = 0.441 g
(b) KOH is in excess; moles of the excess KOH = 0.0014 mole
8.
Molarity of H2SO4 = 0.155 M
Exercises #4:
1.
Oxidation numbers of each atom in componds:
(a) CaCO3: Ca = +2; C = +4; O = -2;
(d) KClO3: K = +1; Cl = +5; O = -2;
(b) Cr2O72-: Cr = +6; O = -2;
(d) MnO4-: Mn = +7; O = -2;
(c) HNO3: H = +1; N = +5; O = -2;
2.
Determine whether each of the following reactions is a redox or non-redox? Explain.
(a) CaCO3(s)  CaO(s) + CO2(g);
Non-redox; none of the atom changes oxidation number;
(b) 2 KClO3(s)  2 KCl(s) + 3 O2(g);
Redox; Cl is reduced (+5 to -1), and O is oxidized (-2 to 0)
(c) (NH4)2Cr2O7(s)  Cr2O3(s) + N2(g) + 4H2O(g); Redox;
N is oxidized (-3 to 0); Cr is reduced (+6 to +3)
(d) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq)  Mn2+(aq) + 5Fe3+(aq) + 4H2O(l):
Redox
Mn is reduced (+7 to +2); Fe is oxidized (+2 to +3)
(e) CaO(s) + SO2(g)  CaSO3(s); Non-redox; no change is oxidation number
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Chem 1A
Keys to Chapter 4 Exercises
3.
Name:_________________________
(a) Net ionic equation:
2 MnO4–(aq) + 5 HSO3–(aq) + 6 H+(aq)  2 Mn2+(aq) + 5 HSO4–(aq) + 3 H2O(aq)
Molecular equation:
2KMnO4(aq) + 5NaHSO3(aq) + 3H2SO4(aq)  2MnSO4(aq) + 5NaHSO4(aq) + 3H2O(aq) + K2SO4(aq)
(b) Cr2O72–(aq) + 3 H2O2(aq) + 8 H+(aq)  2 Cr3+(aq) + 7 H2O(l) + 3 O2(g)
4.
(a) 2 Cr(OH)4-(aq) + 3 H2O2(aq) + 2 OH–(aq)  2 CrO42–(aq) + 8 H2O(l)
–
–
(b) 3 Br2(aq) + 6 OH-(aq)  BrO3 (aq) + 5 Br (aq) + 3 H2O(l)
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